Problem
Numbers can be regarded as product of its factors. For example,
8 = 2 x 2 x 2;
= 2 x 4.
Write a function that takes an integer n and return all possible combinations of its factors.
**Note: **
- Each combination's factors must be sorted ascending, for example: The factors of 2 and 6 is
[2, 6], not[6, 2]. - You may assume that n is always positive.
- Factors should be greater than 1 and less than n.
Examples:
input: 1
output:
[]
input: 37
output:
[]
input: 12
output:
[
[2, 6],
[2, 2, 3],
[3, 4]
]
input: 32
output:
[
[2, 16],
[2, 2, 8],
[2, 2, 2, 4],
[2, 2, 2, 2, 2],
[2, 4, 4],
[4, 8]
]
Solution
DFS的思想。想枚舉出所有的factors，然后dfs依次對(duì)每個(gè)因子試除，如果余數(shù)為0，則可視為一個(gè)結(jié)果。并把它放入最終結(jié)果。這里要注意如果 d = n / factors，并且 d >= factors，才把它放入結(jié)果，首先保證了sorted，另外如果d < factors，說(shuō)明之前肯定有 factors' = d 的情況已經(jīng)做過(guò)了，沒(méi)有必要再做一次。
class Solution {
private:
vector<int> factors;
public:
void solve(int dep, int maxDep, int n, vector<int> &ans, vector<vector<int>> &ret) {
for(int i = dep; i < maxDep; i++) {
int mod = n % factors[i];
if (mod == 0) {
int d = n / factors[i];
ans.push_back(factors[i]);
if (d >= factors[i]) {
vector<int> a(ans);
a.push_back(d);
ret.push_back(a);
if (d > factors[i]) {
solve(i, maxDep, d, ans, ret);
}
}
ans.pop_back();
}
}
}
vector<vector<int>> getFactors(int n) {
for(int i = 2; i < n; i++) {
if (n % i == 0) {
factors.push_back(i);
}
}
vector<vector<int>> ret;
vector<int> ans;
if (factors.size() != 0) {
solve(0, factors.size(), n, ans, ret);
}
return ret;
}
};
